# There's a question that asks for the 2nd derivative of y−2x−3xy=2

Abdiel Mays 2022-11-20 Answered
Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get:
$\frac{12+18y}{\left(1-3x{\right)}^{2}}$
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:
$\frac{48}{\left(1-3x{\right)}^{3}}$
I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?
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## Answers (2)

mainzollbtt
Answered 2022-11-21 Author has 13 answers
Sure you can in this case
$y=\frac{2\left(x+1\right)}{1-3x}$
and then
${y}^{\prime }=\frac{8}{\left(1-3x{\right)}^{2}}$
and finally
${y}^{″}=\frac{48}{\left(1-3x{\right)}^{3}}$
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Elliana Molina
Answered 2022-11-22 Author has 3 answers
We differentiate once,
${y}^{\prime }-2-3y-3x{y}^{\prime }=0$
and twice,
${y}^{″}-3{y}^{\prime }-3{y}^{\prime }-3x{y}^{″}=0.$
Now we can eliminate ${y}^{\prime }$, using
$\left(1-3x\right){y}^{\prime }=3y+2$
and
$\left(1-3x\right){y}^{″}=6{y}^{\prime }=6\frac{3y+2}{1-3x}.$
You can also eliminate y using the initial equation.
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